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Math Help - having trouble with matrix!

  1. #1
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    Exclamation having trouble with matrix!

    I'm trying to work through this prob but not getting anywhere. Also can't find any example in my text or on line. I don't want to give up, can someone explain where to start? OK I'm supposed to put this matrix into echelon form:
    r1- 1 a b+c
    r2- 1 b a+c
    r3- 1 c b+a
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  2. #2
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    Quote Originally Posted by hatsula View Post
    I'm trying to work through this prob but not getting anywhere. Also can't find any example in my text or on line. I don't want to give up, can someone explain where to start? OK I'm supposed to put this matrix into echelon form:
    r1- 1 a b+c
    r2- 1 b a+c
    r3- 1 c b+a
    I don't understand what you have there. Can you post the question verbatim?
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  3. #3
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    Quote Originally Posted by hatsula View Post
    I'm trying to work through this prob but not getting anywhere. Also can't find any example in my text or on line. I don't want to give up, can someone explain where to start? OK I'm supposed to put this matrix into echelon form:
    r1- 1 a b+c
    r2- 1 b a+c
    r3- 1 c b+a
    After staring at this for some time, I believe you meant

    \begin{pmatrix}1 & a & b+c\\ 1 & b & a+c \\ 1 & c & b+a\end{pmatrix}

    Just to clarify, do you need row echelon form, or reduced row echelon form?

    Did your professor not explain the "rules" for manipulating matrices to get echelon form? The rules are to use elementary row operations, of which there are three, (1) switch rows, (2) multiply a row by a nonzero constant, (3) add a multiple of one row to another row. These are called row switching, row multiplication, and row addition, respectively, and a more formal treatment is given here.
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  4. #4
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    yes

    Quote Originally Posted by undefined View Post
    After staring at this for some time, I believe you meant

    \begin{pmatrix}1 & a & b+c\\ 1 & b & a+c \\ 1 & c & b+a\end{pmatrix}

    Just to clarify, do you need row echelon form, or reduced row echelon form?

    Did your professor not explain the "rules" for manipulating matrices to get echelon form? The rules are to use elementary row operations, of which there are three, (1) switch rows, (2) multiply a row by a nonzero constant, (3) add a multiple of one row to another row. These are called row switching, row multiplication, and row addition, respectively, and a more formal treatment is given here.
    OK Yes That is what my problem looks like, and weve been doing matix for a couple weeks so I know the rules, I need to put this into row echelon not reduced. Can I just multiply certain rows by fractions like Bx1/B to get 1? Is that all I have to do? I can't beleive I just thought of that, I've been looking at this problem for a day and a half! Am i going in the right direction? Finals are like in two weeks and I really want to get this stuff be fore school ends!
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  5. #5
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    You need to obtain a 0 in a_{21} position before you attempt to get a 1 in the a_{22} position and so on for row 3.
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    Quote Originally Posted by undefined View Post
    After staring at this for some time, I believe you meant

    \begin{pmatrix}1 & a & b+c\\ 1 & b & a+c \\ 1 & c & b+a\end{pmatrix}

    Just to clarify, do you need row echelon form, or reduced row echelon form?

    Did your professor not explain the "rules" for manipulating matrices to get echelon form? The rules are to use elementary row operations, of which there are three, (1) switch rows, (2) multiply a row by a nonzero constant, (3) add a multiple of one row to another row. These are called row switching, row multiplication, and row addition, respectively, and a more formal treatment is given here.
    Matbe I've figured this out:
    I did r1-r2=r2 = 0 a-c c-a
    r1-r3=r3 = 0 a-b b-a

    then I multiplied r2 times 1/(a-c) and r3 times 1/(a-b)

    so now my matrix looks like this: 1 a (b+c)

    0 1 -1

    0 1 -1

    then I subtracted r3-r2= r3 = 0 0 0 and there I have my echelon form correct? now the next question in my book says find the determinant. But since I have a whole row of zeros doestn't that mean that the determinant is zero?
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  7. #7
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    That is it. It also means the matrix is singular(det=0), the column vectors are linear dependent, the rank is 2, and the nullity is 1.
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    That is it. It also means the matrix is singular(det=0), the column vectors are linear dependent, the rank is 2, and the nullity is 1.
    awesome ur a doll! thank U!!
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